# C# Program to Find the Number Occurring Odd Number of Times

Write a C# program for a given array of positive integers. All numbers occur an even number of times except one number which occurs an odd number of times. Find the number in O(n) time & constant space.

Examples :

Input: arr = {1, 2, 3, 2, 3, 1, 3}
Output : 3

Input: arr = {5, 7, 2, 7, 5, 2, 5}
Output: 5

## C# Program to Find the Number Occurring Odd Number of Times using Nested Loop:

A Simple Solution is to run two nested loops. The outer loop picks all elements one by one and the inner loop counts the number of occurrences of the element picked by the outer loop.

Below is the implementation of the brute force approach :

## C#

 `// C# program to find the element ` `// occurring odd number of times` `using` `System;`   `class` `GFG` `{` `    ``// Function to find the element ` `    ``// occurring odd number of times` `    ``static` `int` `getOddOccurrence(``int` `[]arr, ``int` `arr_size)` `    ``{` `        ``for` `(``int` `i = 0; i < arr_size; i++) {` `            ``int` `count = 0;` `            `  `            ``for` `(``int` `j = 0; j < arr_size; j++) {` `                ``if` `(arr[i] == arr[j])` `                    ``count++;` `            ``}` `            ``if` `(count % 2 != 0)` `                ``return` `arr[i];` `        ``}` `        ``return` `-1;` `    ``}` `    `  `    ``// Driver code ` `    ``public` `static` `void` `Main()` `    ``{` `        ``int` `[]arr = { 2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2 };` `        ``int` `n = arr.Length;` `        ``Console.Write(getOddOccurrence(arr, n));` `    ``}` `}`   `// This code is contributed by Sam007`

Output

```5
```

Time Complexity: O(n^2)
Auxiliary Space: O(1)

## C# Program to Find the Number Occurring Odd Number of Times using Hashing:

A Better Solution is to use Hashing. Use array elements as a key and their counts as values. Create an empty hash table. One by one traverse the given array elements and store counts.

Below is the implementation of the above approach:

## C#

 `// C# program to find the element occurring odd ` `// number of times` `using` `System;` `using` `System.Collections.Generic; `   `public` `class` `OddOccurrence ` `{` `    ``// function to find the element occurring odd` `    ``// number of times` `    ``static` `int` `getOddOccurrence(``int` `[]arr, ``int` `n)` `    ``{` `        ``Dictionary<``int``,``int``> hmap = ``new` `Dictionary<``int``,``int``>();` `        `  `        ``// Putting all elements into the HashMap` `        ``for``(``int` `i = 0; i < n; i++)` `        ``{` `            ``if``(hmap.ContainsKey(arr[i]))` `            ``{` `                ``int` `val = hmap[arr[i]];` `                        `  `                ``// If array element is already present then` `                ``// increase the count of that element.` `                ``hmap.Remove(arr[i]);` `                ``hmap.Add(arr[i], val + 1); ` `            ``}` `            ``else` `                `  `                ``// if array element is not present then put` `                ``// element into the HashMap and initialize ` `                ``// the count to one.` `                ``hmap.Add(arr[i], 1); ` `        ``}`   `        ``// Checking for odd occurrence of each element present` `        ``// in the HashMap ` `        ``foreach``(KeyValuePair<``int``, ``int``> entry ``in` `hmap)` `        ``{` `            ``if``(entry.Value % 2 != 0)` `            ``{` `                ``return` `entry.Key;` `            ``}` `        ``}` `        ``return` `-1;` `    ``}` `        `  `    ``// Driver code ` `    ``public` `static` `void` `Main(String[] args) ` `    ``{` `        ``int` `[]arr = ``new` `int``[]{2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2};` `        ``int` `n = arr.Length;` `        ``Console.WriteLine(getOddOccurrence(arr, n));` `    ``}` `}`   `// This code is contributed by Princi Singh`

Output

```5
```

Time Complexity: O(n)
Auxiliary Space: O(n)

## C# Program to Find the Number Occurring Odd Number of Times using Bit Manipulation:

The Best Solution is to do bitwise XOR of all the elements. XOR of all elements gives us odd occurring elements.

`Here ^ is the XOR operators;Note :x^0 = xx^y=y^x (Commutative property holds)(x^y)^z = x^(y^z) (Distributive property holds)x^x=0`

Below is the implementation of the above approach.

## C#

 `// C# program to find the element` `// occurring odd number of times` `using` `System;`   `class` `GFG` `{` `    ``// Function to find the element ` `    ``// occurring odd number of times` `    ``static` `int` `getOddOccurrence(``int` `[]arr, ``int` `arr_size)` `    ``{` `        ``int` `res = 0;` `        ``for` `(``int` `i = 0; i < arr_size; i++) ` `        ``{` `            ``res = res ^ arr[i];` `        ``}` `        ``return` `res;` `    ``}` `    `  `    ``// Driver code` `    ``public` `static` `void` `Main()` `    ``{` `        ``int` `[]arr = { 2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2 };` `        ``int` `n = arr.Length;` `        ``Console.Write(getOddOccurrence(arr, n));` `    ``}` `}`   `// This code is contributed by Sam007`

Output

```5
```

Time Complexity: O(n)
Auxiliary Space: O(1)

Please refer complete article on Find the Number Occurring Odd Number of Times for more details!

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